Problem 34 - General Chemistry: Principles and Modern Applications

Question

What electron transition in a hydrogen atom, ending in the orbit \(n=3,\) will produce light of wavelength \(1090 \mathrm{nm} ?\)

Step-by-Step Solution

Answer

The electron transition which will produce light of 1090 nm when ending in orbit n=3, begins in orbit \(n=2\).

Step 1: Convert the Wavelength

Firstly, convert the given wavelength into meters because the Rydberg constant has units of \(m^{-1}\). The wavelength, \(λ\), is given as 1090 nm, which is equal to \(1.090 \times 10^{-6} m\) since 1 nm equals \(1 \times 10^{-9} m\).

Step 2: Substitute in the Rydberg Formula

Next, substitute the converted wavelength and given values into the Rydberg formula. For hydrogen, \(R_H = 1.097 \times 10^7 m^{-1}\) and \(n_2=3\). So, \(1/(1.090 \times 10^{-6} m) = 1.097 \times 10^7 [(1/n_1^2) - (1/3^2)]\).

Step 3: Solve the Equation

Rearrange the equation to solve for \(n_1\), square root the result to find \(n_1\). After calculation, you will find \(n_1=2\).

Key Concepts

Rydberg formula

When dealing with electron transitions in hydrogen atoms, the Rydberg formula is a crucial tool. It helps calculate the wavelengths of light associated with electron transitions between energy levels. The formula is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where:

• \(\lambda\) is the wavelength of emitted light.
• \(R_H\) is the Rydberg constant, which is approximately \(1.097 \times 10^7 \ m^{-1}\).
• \(n_1\) and \(n_2\) are the principal quantum numbers of the initial and final energy levels, with \(n_2 > n_1\).

Using this formula, you can calculate the specific wavelengths of light that correspond to electron transitions, revealing the unique spectral lines of the hydrogen atom. This approach not only helps in identifying electron transitions but also deepens your understanding of atomic spectroscopy.

Atomic orbitals

Atomic orbitals are regions around the nucleus where electrons are likely to be found. Each orbital corresponds to a different energy level within an atom, typically denoted by the principal quantum number, \(n\). In the context of the hydrogen atom:

• \(n=1\) represents the first energy level and the closest orbital to the nucleus.
• Higher \(n\) values correspond to orbitals that are farther from the nucleus and have higher energy.

When electrons transition between these orbitals, they absorb or emit specific amounts of energy. In the exercise, an electron transition from an orbital with \(n_1=2\) to \(n_2=3\) in a hydrogen atom results in light emission as the electron moves to a lower energy state. Understanding atomic orbitals is key to grasping how electronic transitions cause the emission or absorption of light, thus forming the basis for spectra and chemical bonding.

Wavelength conversion

Converting wavelengths is a fundamental step in applying the Rydberg formula. Often, wavelengths are given in nanometers (nm) but need to be converted into meters (m) for consistency with the units of the Rydberg constant. Here’s how you can do it:

• Recognize that 1 nm is equal to \(1 \times 10^{-9} m\).
• To convert a wavelength from nanometers to meters, multiply by \(1 \times 10^{-9}\).

In the exercise, the given wavelength of 1090 nm converts to 1.090 x 10^-6 m. This conversion ensures that when the wavelength is used in the Rydberg formula, all units remain consistent. Mastering wavelength conversion is essential for analyzing spectral lines and electronic transitions across different systems, reinforcing the importance of working within a unified unit system.